For all $x,y∈\Bbb{R}$ define that $ x\equiv y$ if$ x^2=y^2$

Question

For all $x,y\in\Bbb{R}$ define that $x\equiv y$ if $x^2=y^2$ . Then $\equiv$ is an equivalence relation on $\Bbb{R}$ , there are infinitely many equivalence classes, one of them consists of one element and the rest consist of two elements.

Solution

True. To show that $\equiv$ is reflexive we need to show that $\forall x\in\mathbb{R} :x\equiv x$. Let $x\in\mathbb{R} , x\equiv x \mbox{ if } x^{2}=x^{2}$, which is obvious.

$[x] ={y\in\mathbb{R} |x\equiv y} =[0]={y\in\mathbb{R}|0\equiv y}={0}$. Hence $y^{2} =0^{2} =0$ which implies that $y=0$.

Can anyone please give me correct answer to this question.

Answer 1

$$x\equiv y \iff x=y \lor x=-y$$ Classes are $\{0\}$ and $\{-x, x\}$ for all $x>0$.

Answer 2

Proof of symmetry and transitivity:

Symmetry:
Let $x,y\in\Bbb{R}$ with $x\equiv y$, so we have $x²=y²$ and trivially we get $y²=x²$, hence $y\equiv x$.

Transitivity:
Let $x,y,z\in\Bbb{R}$ with $x\equiv y$ and $y\equiv z$. We now show that $x\equiv z$: Since $x²=y²$ and $y²=z²$ we get by transitivity of equality $x²=z²$ and therefore $x\equiv z$.